Question

friends solve this equation.​

Answer 1

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Answer 2

Answer:

$cos^{-1}(\frac{x-\frac{1}{x}}{x+\frac{1}{x}})=\pi-2\:tan^{-1}x$

Step-by-step explanation:

Formula used:

$cos2A=\frac{1-tan^2A}{1+tan^2A}$

$cos^{-1}(-x)=\pi-cos^{-1}x$

Now,

$cos^{-1}(\frac{x-\frac{1}{x}}{x+\frac{1}{x}})$

$=cos^{-1}(\frac{\frac{x^2-1}{x}}{\frac{x^2+1}{x}})$

$=cos^{-1}(\frac{x^2-1}{x^2+1})$

$=cos^{-1}[-(\frac{1-x^2}{1+x^2})]$

................................................

Take

$x=\tan\theta$

then $\theta=tan^{-1}x$

...............................................

$=cos^{-1}[-(\frac{1-\tan^2\theta}{1+\tan^2\theta})]$

$=cos^{-1}[-(cos2\theta)]$

$=\pi-cos^{-1}(cos2\theta)$

$=\pi-2\theta$

$=\pi-2\:tan^{-1}x$